• # question_answer 45) The adjacent figure shows the position graph of one dimensional motion of a particle of mass 4 kg. The impulse at $t=0\,s$ and $t=4$is given respectively as: A)  $0,0$B)  $0,-3\,kg\,m{{s}^{-1}}$C)  $+3\,kg\,m{{s}^{-1}},0$D)  $+3\,kg\,m{{s}^{-1}},-3kg.m{{s}^{-1}}$

$t<0,\,{{v}_{i}}=0$and $t>0,\,{{v}_{f}}=\frac{3}{4}m/s$ $\therefore$ Impulse $=m({{v}_{f}}-{{v}_{i}})=4\left( \frac{3}{4}-0 \right)$                 $=3\,kg\,m{{s}^{-1}}$ Little before 4 second $v=\frac{3}{4}m/s,$ little after 4 second, velocity becomes zero Therefore, Impulse $=m({{v}_{f}}-{{v}_{i}})=4\left( 0-\frac{3}{4} \right)=-3kg\,m{{s}^{-1}}$